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Mathematics · 2026 · 3 marks
Differential EquationsMethods of Solving First Order, First Degree Differential Equations3 marksApplyshort answer
CBSE 2026 · Region 1 · Set 2 · Q27
(a) Find the general solution of the differential equation: y^2 d x +( x^2- x y + y^2) dy =0
(a)
Find the general solution of the differential equation : \[\mathrm{y}^{2} \mathrm{~d} \mathrm{x}+\left(\mathrm{x}^{2}-\mathrm{x} \mathrm{y}+\mathrm{y}^{2}\right) \mathrm{dy}=0 \]
OR
(b)
Find the particular solution of the differential equation $\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y} \tan \mathrm{x}$, given that $\displaystyle \mathrm{y}=2$ if $\displaystyle \mathrm{x}=0$.
Marking-scheme solution
Given differential equation is $\displaystyle \dfrac{dx}{dy} = -\left(\dfrac{x}{\mathrm{y}}\right)^2 + \dfrac{x}{\mathrm{y}} - 1$
Put $\displaystyle x = v\mathrm{y} \Rightarrow \dfrac{dx}{dy} = v + \mathrm{y}\dfrac{dv}{dy}$
The given differential equation becomes $\displaystyle \mathrm{y}\dfrac{dv}{dy} = -1 - v^2$
$\displaystyle \Rightarrow \dfrac{dv}{1+v^2} = -\dfrac{dy}{\mathrm{y}}$
$\displaystyle \Rightarrow \int \dfrac{dv}{1+v^2} = -\int \dfrac{dy}{\mathrm{y}}$
$\displaystyle \Rightarrow \tan^{-1}v = -\log|\mathrm{y}| + C$
$\displaystyle \Rightarrow \tan^{-1}\dfrac{x}{\mathrm{y}} = -\log|\mathrm{y}| + C$
Given differential equation is $\displaystyle \dfrac{dy}{\mathrm{y}} = \tan x\,dx$
$\displaystyle \Rightarrow \int \dfrac{dy}{\mathrm{y}} = \int \tan x\,dx$
$\displaystyle \Rightarrow \log|\mathrm{y}| = \log|\sec x| + \log C,$
$\displaystyle \Rightarrow \log\left|\dfrac{\mathrm{y}}{\sec x}\right| = \log C$
$\displaystyle \Rightarrow \dfrac{\mathrm{y}}{\sec x} = C$
$\displaystyle \Rightarrow \mathrm{y} = C\sec x$
When $\displaystyle x = 0, \mathrm{y} = 2 \Rightarrow C = 2$
Hence, the required particular solution is $\displaystyle \mathrm{y} = 2\sec x$
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.